# $\triangle ABC$ has $AC=BC$, and $\angle ACB=96^\circ$. $D$ is a point such that $\angle DAB=18^\circ, \angle DBA=30^\circ$. What is $\angle ACD$?

$\triangle ABC$ has $AC=BC$, and $\angle ACB=96^\circ$. $D$ is a point in $\triangle ABC$, such that $\angle DAB=18^\circ, \angle DBA=30^\circ$. What is $\angle ACD$?

My attempt:

$$\angle ABC=\angle BAC=\frac{(180^\circ-96^\circ)}{2}=42^\circ.$$

$$\angle ADB=180^\circ-18^\circ-30^\circ=132^\circ.$$

From here onwards, I have no idea how to carry on. Can anyone help?

When I see that $\angle CAB$ is partitioned as $24^\circ$ and $18^\circ$, what angle is left to make an $60^\circ$ angle (which is for making an equilateral triangle, which gives us more equal sides)? After doing that, we see that $\angle ECB = 36^\circ$ and this makes $\angle ABE = 30^\circ$. Now, notice that $\Delta ADB$ is congruent to $\Delta AEB$. So we have $|AD| = |AE| = |AC|$. So the answer is $78^\circ$.

By law of sines we obtain: $$\frac{\sin(96^{\circ}-x)}{DB}=\frac{\sin12^{\circ}}{CD}$$ and $$\frac{\sin{x}}{AD}=\frac{\sin24^{\circ}}{CD},$$ which gives $$\frac{\sin(96^{\circ}-x)}{\sin{x}}\cdot\frac{AD}{DB}=\frac{\sin12^{\circ}}{\sin24^{\circ}}$$ or $$\frac{\sin(96^{\circ}-x)}{\sin{x}}\cdot\frac{\sin30^{\circ}}{\sin18^{\circ}}=\frac{\sin12^{\circ}}{\sin24^{\circ}}$$ or $$\sin96^{\circ}\cot{x}-\cos96^{\circ}=\frac{\sin18^{\circ}}{\cos12^{\circ}}$$ or $$\cos6^{\circ}\cot{x}=\frac{\sin18^{\circ}-\sin6^{\circ}\cos12^{\circ}}{\cos12^{\circ}}$$ or $$\cos6^{\circ}\cot{x}=\frac{\sin18^{\circ}-\frac{1}{2}(\sin18^{\circ}-\sin6^{\circ})}{\cos12^{\circ}}$$ or $$\cos6^{\circ}\cot{x}=\frac{\sin12^{\circ}\cos6^{\circ}}{\cos12^{\circ}},$$ which gives $x=78^{\circ}.$

• Yes, but this is an olympiad question. Any idea how to get x=78 from $$\sin96^{\circ}\cot{x}-\cos96^{\circ}=\frac{\sin18^{\circ}}{\cos12^{\circ}},$$ without the use of a calculator? Jan 20, 2018 at 11:35
• @XCoderX Note that: $$\frac{\sin(96^\circ - x)}{\sin x} \frac{1}{2\sin 18^\circ}=\frac1{2\cos 12^\circ}$$ $$\implies \frac{\sin(96^\circ-x)}{\sin x} =\frac{\sin 18^\circ}{\sin 78^\circ}$$
– user371838
Jan 20, 2018 at 11:39

Rotate $C$ for $60^{\circ}$ around $A$ (we get new point $E$). Note that $E$ and $B$ are on different side of line $AC$.

Then $E$, $D$ and $B$ are colinear (calculate the angle $∠EBC$ (look at triangle $EBC$) and the angle $∠DBC$ (look at the angle $∠ABC$))

so $\angle ADE = 48 = \angle AED$ so $ADE$ is isosceles and so is $ACD$. Thus $\angle ACD = 78$.

Take $$O$$ the circumcenter of $$\triangle ABD$$, so $$AO=OD=OB=AD$$ and $$\angle BAO=\angle OBA=60^\circ-\angle BAD=42^\circ$$, thus $$\triangle AOB\cong\triangle BCA$$, implying $$OA=AC=AD$$, thence $$\angle ADC=\angle ACD=90^\circ-12^\circ=78^\circ$$.